Liquid ammonia is produced at high temperatures and under great pressure in a tank by

12 answers
Question:

Liquid ammonia is produced at high temperatures and under great pressure in a tank by passing a mixture of nitrogen gas and hydrogen gas over an iron catalyst. the reaction is represented by the following equation. n2(g) + 3h2(g) → 2nh3(g) changing all but one experimental condition will affect the amount of ammonia produced. that condition is a) increasing the concentration of both reactants b) changing the temperature within the tank c) decreasing the pressure within the tank. d) increasing only the amount of nitrogen present.

Answers

a) [tex]\mu_{k}=0.367[/tex] b) [tex]V_{B}=3.98m/s[/tex]

Explanation:

In order to start solving this problem, we must begin by drawing a diagram of the situation (see attached picture).

On the diagram, we can identify three important points:

Point A=Starting point

Point B=Point where the block detaches from the spring.

Point C= Point where the block stops.

So in order to solve part a, we need to analyze the movement from point A to point C. We can do that by building an energy balance equation.

On point A there will only be potential energy while on point C there is no energy at all, since it was all wasted by the work done by the friction, so the equation will look like this:

[tex]U_{s}=fx[/tex]

We know that the friction is defined to be the coefficient of kinetic friction times the normal force, so we can substitute that in our equation:

[tex]U_{s}=N\mu_{k}x[/tex]

we can now solv this for the coefficient of kinetic friction so we get:

[tex]\mu_{k}=\frac{U_{s}}{Nx}[/tex]

if we do a free body diagram of the block and do a sum of forces in the y-direction we get that:

[tex]\sum F_{x}=0[/tex]

[tex]N-W=0[/tex]

so

N=W

or

N=mg

we can substitute this into our coefficient of kinetic friction equation, so we get:

[tex]\mu_{k}=\frac{U_{s}}{mgx}[/tex]

we know that the potential energy of a spring is given by:

[tex]U_{s}=\frac{1}{2}kd^{2}[/tex]

so we can substitute that also into our equation, so we get:

[tex]\mu_{k}=\frac{kd^{2}}{2mgx}[/tex]

so now we can substitute data:

[tex]\mu_{k}=\frac{(100N/m)(0.3m)^{2}}{2(0.5kg)(9.81m/s^{2})(2.5m)}[/tex]

which yields:

[tex]\mu_{k}=0.367[/tex]

Once we have solved for the coefficient of kinetic friction, we can start solving part b:

b)

If we do an energy balance from point A to point B, we can see that in point A there will only be potential energy, while at point B there will be kinetic energy plut the energy lost due to the work done by the friction, so our equation looks like this:

[tex]U_{sA}=K_{B}+fd[/tex]

We can solve this equation for the kinetic energy, so we get:

[tex]K_{B}=U_{sA}-fd[/tex]

we know that the kinetic energy is defined to be one half of the mass times the square of the velocity. We also know what the potential energy of a spring is so we can substitute that into our equation, so we get:

[tex]\frac{1}{2}mV_{B}^{2}=\frac{1}{2}kd^{2}-fd[/tex]

we can now multiply both sides of the equation by 2 so we get:

[tex]mV_{B}^{2}=kd^{2}-2fd[/tex]

so now we can solve for the velocity, so we get:

[tex]V_{B}=\sqrt{\frac{kd^{2}-2fd}{m}}[/tex]

from the analysis done on the previous step, we know that the force of friction is given by the normal times the coefficient of kinetic friction, so we can use that in our equation to get:

[tex]V_{B}=\sqrt{\frac{kd^{2}-2N\mu_{k}d}{m}}[/tex]

and we also know tha tthe normal is given by the weight of the block, so we can also substitute that:

[tex]V_{B}=\sqrt{\frac{kd^{2}-2mg\mu_{k}d}{m}}[/tex]

and now we can substitute all the values provided by the problem:

[tex]V_{B}=\sqrt{\frac{(100N/m)(0.3m)^{2}-2(0.5kg)(9.81m/s^{2})(0.367)(0.3m)}{0.5kg}}[/tex]

Which yields:

[tex]V_{B}=3.98m/s[/tex]


[tex]To measure a coefficient of kinetic friction we can use a spring with a known force constant. a bloc[/tex]

a) W > 0

b) W = 0

c) W < 0

d) W = 0

e) W = 0

f) W = m*g*Sin ϕ*d

g) W = - k*x²/2

Explanation:

a. Is the work done on the box by the force of the push positive, negative, or zero?

The work done on the box by the force of the push is positive

W > 0.

The box has been acted on by a force that tranfers energy to the box, thereby increasing the object's box.

b. Is the work done on the box by the normal force positive, negative, or zero?

The work done on the box by the normal force is zero.

W = 0  since ϕ = 90° ⇒ Cos 90° = 0

c. Is the work done on the box by the force of friction positive, negative, or zero?

The work done on the box by the force of the push is negative

W < 0.

The box has been acted on by a force that has reduced the energy of the box.

d. Is the work done on the box by gravity positive, negative, or zero?

The work done on the box by gravity is zero.

W = 0  since ϕ = 270° ⇒ Cos 270° = 0

e. It's time to rearrange your furniture. You try to push your chest of drawers, with a of weight 3,500 N , across the floor to a spot 5 m away. You push as hard as you can for several seconds until you collapse from exhaustion, but the chest does not move. How much work did you do on the chest?

W = 0 since the displacement is zero (d = 0).

f. A block with mass m slides down a frictionless ramp that is inclined at an angle ϕ above the horizontal. How much work is done on the box by gravity if the box moves a distance d down the slope?

W = m*g*Sin ϕ*d

g. Assume that Earth's orbit around the sun is circular. How much work is done on the Earth by the Sun in one complete orbit? The mass of the Earth is Me, the mass of the Sun is Ms, and the distance from the Sun to the Earth is res

W = 0 since ϕ = 90° ⇒ Cos 90° = 0

h. A rock with mass m is pressed up against a spring with spring constant k, compressing it by x from its relaxed position. How much work is done on the rock by the spring?

W = - k*x²/2

Step-by-step explanation:

[tex]\frac{AB}{BC} =cos 71\\BC=\frac{AB}{cos 71} =\frac{6.3}{cos71} \approx 19.351[/tex]

The annual net cash inflows from the intangible benefits have to be $35,000 to make this a financially acceptable investment

Explanation:

According to the given data we have the following:

required rate of return=14%

Negative net present value=$182,560

Therefore, in order to calculate How large would the annual net cash inflows from the intangible benefits have to be to make this a financially acceptable investment we would have to use the following formula:

Minimum annual cash flows required=Negative net present value/Present value factor at 14% for 10 years

Present value factor at 14% for 10 years=5.216

Therefore, Minimum annual cash flows required=$182,560/5.216

Minimum annual cash flows required=$35,000

The annual net cash inflows from the intangible benefits have to be $35,000 to make this a financially acceptable investment

The answers are BC = 6.3/ cos71 = 19.351 and BC = 6.3 / sin 19 = 19.351

Step-by-step explanation:

cosB = [tex]\frac{base}{hypoteneus}[/tex]

or, cos 71 = [tex]\frac{AB}{BC}[/tex]

or,BC = [tex]\frac{AB}{cos71}[/tex]

= 6.3/ cos71 = 19.351

Here, ∠C = 180 - (90 + 71 ) = 19

sin19 =[tex]\frac{perpendicular}{hypoteneus}[/tex]

or, sin 19 = AB/ BC

or, BC = AB /sin 19

= 6.3 / sin 19 = 19.351

We cannot find BC in terms of tan as tan means perpendicular/ hypotenuse. BC is the hypotenuse and that is got eliminated.

Here is the explanation.

The correct answer is C

Explanation:

I'd say It demonstrates how Miss Fairchild's attitude changes from negative to positive when she recognizes Easton .

Scenes the chair wheels are up the person is rolling backwards and if the wheels were down then the person would go forwards
 

The correct/right answer is D

This question is lacking especially the diagram, well I've read the entire question and seen its diagram and it demonstrates a Coriolis effect. The child that could get the ball would be child c. Well Coriolis effect is the deflection of a free moving object, such as ball, due to a rotation of the merry go round.

b i think it’s a water wheel turning in a stream

the answer would be b i believe

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